Sync LeetCode submission Runtime - 227 ms (60.00%), Memory - 60.2 MB (69.15%)

Author: jimit105

2564-most-profitable-path-in-a-tree/README.md

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+<p>There is an undirected tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>, rooted at node <code>0</code>. You are given a 2D integer array <code>edges</code> of length <code>n - 1</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>
+
+<p>At every node <code>i</code>, there is a gate. You are also given an array of even integers <code>amount</code>, where <code>amount[i]</code> represents:</p>
+
+<ul>
+	<li>the price needed to open the gate at node <code>i</code>, if <code>amount[i]</code> is negative, or,</li>
+	<li>the cash reward obtained on opening the gate at node <code>i</code>, otherwise.</li>
+</ul>
+
+<p>The game goes on as follows:</p>
+
+<ul>
+	<li>Initially, Alice is at node <code>0</code> and Bob is at node <code>bob</code>.</li>
+	<li>At every second, Alice and Bob <b>each</b> move to an adjacent node. Alice moves towards some <strong>leaf node</strong>, while Bob moves towards node <code>0</code>.</li>
+	<li>For <strong>every</strong> node along their path, Alice and Bob either spend money to open the gate at that node, or accept the reward. Note that:
+	<ul>
+		<li>If the gate is <strong>already open</strong>, no price will be required, nor will there be any cash reward.</li>
+		<li>If Alice and Bob reach the node <strong>simultaneously</strong>, they share the price/reward for opening the gate there. In other words, if the price to open the gate is <code>c</code>, then both Alice and Bob pay&nbsp;<code>c / 2</code> each. Similarly, if the reward at the gate is <code>c</code>, both of them receive <code>c / 2</code> each.</li>
+	</ul>
+	</li>
+	<li>If Alice reaches a leaf node, she stops moving. Similarly, if Bob reaches node <code>0</code>, he stops moving. Note that these events are <strong>independent</strong> of each other.</li>
+</ul>
+
+<p>Return<em> the <strong>maximum</strong> net income Alice can have if she travels towards the optimal leaf node.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/10/29/eg1.png" style="width: 275px; height: 275px;" />
+<pre>
+<strong>Input:</strong> edges = [[0,1],[1,2],[1,3],[3,4]], bob = 3, amount = [-2,4,2,-4,6]
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> 
+The above diagram represents the given tree. The game goes as follows:
+- Alice is initially on node 0, Bob on node 3. They open the gates of their respective nodes.
+  Alice&#39;s net income is now -2.
+- Both Alice and Bob move to node 1. 
+&nbsp; Since they reach here simultaneously, they open the gate together and share the reward.
+&nbsp; Alice&#39;s net income becomes -2 + (4 / 2) = 0.
+- Alice moves on to node 3. Since Bob already opened its gate, Alice&#39;s income remains unchanged.
+&nbsp; Bob moves on to node 0, and stops moving.
+- Alice moves on to node 4 and opens the gate there. Her net income becomes 0 + 6 = 6.
+Now, neither Alice nor Bob can make any further moves, and the game ends.
+It is not possible for Alice to get a higher net income.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/10/29/eg2.png" style="width: 250px; height: 78px;" />
+<pre>
+<strong>Input:</strong> edges = [[0,1]], bob = 1, amount = [-7280,2350]
+<strong>Output:</strong> -7280
+<strong>Explanation:</strong> 
+Alice follows the path 0-&gt;1 whereas Bob follows the path 1-&gt;0.
+Thus, Alice opens the gate at node 0 only. Hence, her net income is -7280. 
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>edges.length == n - 1</code></li>
+	<li><code>edges[i].length == 2</code></li>
+	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
+	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
+	<li><code>edges</code> represents a valid tree.</li>
+	<li><code>1 &lt;= bob &lt; n</code></li>
+	<li><code>amount.length == n</code></li>
+	<li><code>amount[i]</code> is an <strong>even</strong> integer in the range <code>[-10<sup>4</sup>, 10<sup>4</sup>]</code>.</li>
+</ul>

2564-most-profitable-path-in-a-tree/solution.py

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+# Approach 1: Depth-First Search and Breadth-First Search
+
+# Time: O(n)
+# Space: O(n)
+
+from collections import deque
+
+class Solution:
+    def __init__(self):
+        self.bob_path = {}
+        self.visited = []
+        self.tree = []
+
+
+    def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
+        n = len(amount)
+        max_income = float('-inf')
+        self.tree = [[] for _ in range(n)]
+        self.bob_path = {}
+        self.visited = [False] * n
+        node_queue = deque([(0, 0, 0)])
+
+        for edge in edges:
+            self.tree[edge[0]].append(edge[1])
+            self.tree[edge[1]].append(edge[0])
+        
+        self.find_bob_path(bob, 0)
+
+        # BFS
+        self.visited = [False] * n
+        while node_queue:
+            source_node, time, income = node_queue.popleft()
+
+            # Alice reaches node first
+            if source_node not in self.bob_path or time < self.bob_path[source_node]:
+                income += amount[source_node]
+            # Reach the node at the same time
+            elif time == self.bob_path[source_node]:
+                income += amount[source_node] // 2
+
+            # Update max value if current node is a new leaf
+            if len(self.tree[source_node]) == 1 and source_node != 0:
+                max_income = max(max_income, income)
+
+            # Explore adjacent unvisited vertices
+            for adj_node in self.tree[source_node]:
+                if not self.visited[adj_node]:
+                    node_queue.append((adj_node, time + 1, income))
+
+            # Mark and remove current node
+            self.visited[source_node] = True
+
+        return max_income
+
+    # DFS
+    def find_bob_path(self, source_node, time):
+        self.bob_path[source_node] = time
+        self.visited[source_node] = True
+
+        if source_node == 0:
+            return True
+
+        for adj_node in self.tree[source_node]:
+            if not self.visited[adj_node]:
+                if self.find_bob_path(adj_node, time + 1):
+                    return True
+
+        # If node 0 isn't reached, remove current node from path
+        self.bob_path.pop(source_node, None)
+        return False